Answer
a) $x$ is an eigenvector of $A$.
b) $x$ is an eigenvector of $A$.
c) $x$ is not an eigenvector of $A$.
d) $x$ is not an eigenvector of $A$.
Work Step by Step
Let $A$ be any $n\times n$ matrix. Let $\lambda$ be some scalar for which there exists a non zero vector $x$ such that $Ax=\lambda x$. Then $\lambda$ is called an eigenvalue of $A$ and $x$ is called the corresponding eigenvalue.
Given $A= \left[ {\begin{array}{*{20}{c}}
{-3}&10\\
5&2
\end{array}} \right]$.
a) Given $x = \left[ {\begin{array}{*{20}{c}}
4\\
4
\end{array}} \right]$, then $Ax = \left[ {\begin{array}{*{20}{c}}
{-3}&10\\
5&2
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
4\\
4
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
{28}\\
{28}
\end{array}} \right]$.
Observe that $\left[ {\begin{array}{*{20}{c}}
{28}\\
{28}
\end{array}} \right] =7 \left[ {\begin{array}{*{20}{c}}
4\\
4
\end{array}} \right]$
Therefore, $x$ is an eigenvector of $A$.
b) Given $x = \left[ {\begin{array}{*{20}{c}}
-8\\
4
\end{array}} \right]$, then $Ax = \left[ {\begin{array}{*{20}{c}}
{-3}&10\\
5&2
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
-8\\
4
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
{64}\\
{-32}
\end{array}} \right]$.
Observe that $\left[ {\begin{array}{*{20}{c}}
{64}\\
{-32}
\end{array}} \right] ={-8} \left[ {\begin{array}{*{20}{c}}
{-8}\\
4
\end{array}} \right]$
Therefore, $x$ is an eigenvector of $A$.
c) Given $x = \left[ {\begin{array}{*{20}{c}}
-4\\
8
\end{array}} \right]$, then $Ax = \left[ {\begin{array}{*{20}{c}}
{-3}&10\\
5&2
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
-4\\
8
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
{92}\\
{-4}
\end{array}} \right]$.
Observe that $\left[ {\begin{array}{*{20}{c}}
{92}\\
{-4}
\end{array}} \right] \neq \lambda \left[ {\begin{array}{*{20}{c}}
{-4}\\
8
\end{array}} \right]$ for any scalar $\lambda \in R$
Therefore, $x$ is not an eigenvector of $A$.
d) Given $x = \left[ {\begin{array}{*{20}{c}}
5\\
-3
\end{array}} \right]$, then $Ax = \left[ {\begin{array}{*{20}{c}}
{-3}&10\\
5&2
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
5\\
-3
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
{-45}\\
{19}
\end{array}} \right]$.
Observe that $\left[ {\begin{array}{*{20}{c}}
{-45}\\
{19}
\end{array}} \right] \neq \lambda \left[ {\begin{array}{*{20}{c}}
{5}\\
-3
\end{array}} \right]$ for any scalar $\lambda \in R$
Therefore, $x$ is not an eigenvector of $A$.
